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Let be the integers a,b,c with

[tex]a^2-4b=c^2[/tex]

To be shown that the number [tex]a^2-2b[/tex] can be written as a sum of 2 perfect squares.

Sagot :

[tex]{ a }^{ 2 }-4b={ c }^{ 2 }\\ \\ { a }^{ 2 }-4b+2b={ c }^{ 2 }+2b\\ \\ { a }^{ 2 }-2b={ c }^{ 2 }+{ \left( \sqrt { 2b } \right) }^{ 2 }[/tex]

[tex]b={ 2 }^{ n }\\ \\ n>0[/tex]

(n) is the set of odd natural numbers greater than 0.
crisforp
a^2 - 4b = c^2 <=> a^2 - c^2 = 4b <=> ( a + c )( a - c ) = 4b, cu a, b, c nr. intregi ;
Avem doua posibilitati :
a) a = 2x si b = 2y;
Atunci ( a + c )( a - c ) = 4b <=> x^2 - y^2 = b;
Relatia  a^2 -2b devine 2x^2 +2y^2 = ([tex] \sqrt{2} x[/tex])^2 + ([tex] \sqrt{2} y[/tex])^2, adica suma a doua patrate perfecte ;

b) a = 2x + 1 si b = 2y + 1 ;
In mod analog. obtii ca x^2 - y^2 + x - y = b;
si, dupa ce prelucrezi, ai ca a^2 - 2b = [[tex] \sqrt{2} ( x + 1 / 2 )[/tex]]^2 + [][tex] \sqrt{2} ( y + 1 / 2 )[/tex]^2, adica, suma a doua patrate perfecte .

Bafta!


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